Question

A turbojet–powered aircraft flies at \(V_0=250\ \text{m s}^{-1}\). The (uninstalled) thrust is \(T=60{,}000\ \text{N}\). Fuel heating value is \(44\times 10^{6}\ \text{J kg}^{-1}\); fuel flow is \(\dot m_f=3\ \text{kg s}^{-1}\). The engine thermal efficiency is \(35\%\). Exit pressure equals ambient. What is the propulsion efficiency (in %)?

• A. 32.5
• B. 35.0
• C. 11.4
• D. 92.4

Hint:

Turbojet efficiencies factorize: \(\eta_o=\eta_\text{th}\eta_p\). If thrust and flight speed are known, compute \(\eta_o=(T V_0)/(\dot m_f\mathrm{LHV})\), then divide by the given thermal efficiency to get \(\eta_p\).

Answer 1

The Correct Option is A

Step 1: Overall efficiency from useful power and fuel power.
Useful (propulsive) power delivered to the airframe: \[ P_\text{useful}=T\,V_0=60{,}000\times 250=15.0\times 10^{6}\ \text{W}. \] Fuel chemical power: \[ \dot Q_f=\dot m_f\,\text{LHV}=3\times 44\times 10^{6}=132\times 10^{6}\ \text{W}. \] Overall efficiency: \[ \eta_o=\frac{P_\text{useful}}{\dot Q_f}=\frac{15}{132}\approx 0.1136=11.36\%. \] Step 2: Use \(\eta_o=\eta_\text{th}\,\eta_p\) to get propulsion efficiency.
Given \(\eta_\text{th}=0.35\), \[ \eta_p=\frac{\eta_o}{\eta_\text{th}} =\frac{0.1136}{0.35}\approx 0.3247\approx 32.5\%. \] \[ \boxed{32.5\%} \]